 This lecture is part of an online course on mathematical group theory and will mostly be about Lagrange's theorem. What I'm going to do over the next few lectures is to go over small finite groups in order, introducing various mathematical theorems whenever we need them to classify the groups. You can find a list of such groups in this book by Coxter and Moser Generators and Relations for Discrete Groups. So if you go to table 1.12, let me just magnify it a bit so you can actually see what it says, here's a table of non-Abelian groups of order less than 32. So what we're going to be doing is just going through this table and explaining everything in it. So let's start with groups of order 1. Order 1 is kind of completely trivial to classify. Let me magnify down again. We just get the trivial group with just one element. Order 2 is also not very exciting. The group has two elements, one of which is the identity and one of which isn't, from which it's easy to reduce that the only possibility is a cyclic group of order 2. Order 3, you can give an elementary argument to show that anything is z over 3z, but the argument is starting to get a little bit tiresome. So what we're going to do is to use a theorem called Lagrange's theorem which says that the order of a subgroup h of a group g divides the order of g. The order of a group is its number of elements and is normally denoted by putting a couple of vertical lines on it. So we just say that the order of h divides the order of g. And how does this help with classifying the elements of order 3? Well, a consequence of Lagrange's theorem or a corollary says that the order of an element of g divides the order of big g. Well, what's the order of an element? The order of an element g is the smallest n greater than or equal to 0 such that g to the n is equal to 1. So the order of the identity element is 1 and so on. And the point is if you've got an element of order n, then the elements 1, 2, opt to g, sorry g squared, opt to g to the n minus 1 form a subgroup h of g of order n. So the order of the subgroup generated by g is the same as the order of the element g. And so it must divide g by Lagrange's theorem. Well, how does this help classify the elements of order 3? Well, it can do better than that. We can classify all elements of all groups of order p for p prime. It's quite common to use p as a prime. And the point is the order of an element, so suppose g has ordered p, the order of an element g and g divides p, so is 1 or p. And if it's 1, if and only if g is the identity element, so the group must have elements of order p. So if it has an element of g of order p, then we can just write the group g is equal to all powers of little g. So is cyclic. So we say group is cyclic if it is generated by one element, meaning that it just consists of powers of this element. Now there's an obvious isomorphism of g with z over pz because you met nought to 1, 1 to g, 2 to g squared and so on. So g is isomorphic to the integers modulo p. So we've done the classification of all groups of prime order. In general, the difficulty of classifying groups of some order doesn't depend on the size of the order, but it depends more on the number of prime factors dividing the order. So we see if there's only one prime factor dividing the order, more precisely if the order is just one prime, then it's easy to classify all those groups. Okay, well now we've got to prove Lagrange's theorem. So to motivate this, suppose g acts on set s, then we pick some element s in the set s and look at h, which is the set of elements g with g of s equal s. So it's the elements of the group g that fix the element s. And these obviously form a subgroup. So you can very easily check the product of two elements fixing s fixes s and the inverse of anything fixing s also fixes s. So if we've got any element, any group acting on a set s, we can generate lots of subgroups of g just by looking at elements of the set s and looking at the fixed points. For example, suppose we take an icosahedron and look at the group of its symmetries, we can find a subgroup fixing this vertex and this subgroup will have order five. So if we fix this vertex, then there are five different ways of doing that. So you can see there are quite a few subgroups who order five corresponding to different vertices. Notice that sometimes vertices correspond to the same subgroup of order five. So for instance, these two opposite vertices are the same subgroup, which just consists of the five rotations. Similarly, we get a subgroup of order three by fixing a triangle and looking at the elements which just map this triangle to itself. And you can see there are three of those. And we could fix an edge and that would give us a subgroup of order two because we can, there's a subgroup of order two fixing each edge. When I say fixing the edge, I mean it maps the edge to itself. It might map points of the edge to other points of the edge. Anyway, so this is one way to think about subgroups. They just consist of, they just correspond to points of a set acted on by G. We want to ask converse. Suppose H is a subgroup of G. Can we find a set S and a point S, little s in S, so that H is the fixed subgroup of little s. In other words, the things fixing S. And the answer to this is yes. And we're going to do it by using things called cosets of H. So let's first of all think about what a group actually on a set looks like. So let's pick a set to be say a triangle. And we're just going to group S3 of the six symmetries of a triangle. And if you look at the points of a triangle, you can see you can divide them into various subsets. For instance, if I take this blue point, you can map it to these other blue points under the group G. If I take a red point here, then I can map it into six other red points by acting on the triangle with G. If I take the point in the middle of the triangle, then that's only mapped to itself. And I could also take a point here and that would give me three different points. And each of these things is called an orbit. So these three blue points are called an orbit of G on S, on the set S, where the set S is now going to be all points in this triangle. And you can see that any set acted on by G splits up into orbits. That's not very difficult to see. You just say two points are equivalent if there's an element of G mapping one to the other. And you can easily see that this is actually the equivalence relation using the group axioms. And so the set splits up into equivalence classes, which are the orbits. And set S acted on by the action of G on S is called transitive if there is only one orbit. So the action of G on these three blue points is a transitive action on the blue points. And the action of G on the six red points is a transitive action on the six red points. Since you can split up any action of a group on into orbits in which this transitive, you may as well for most purposes just consider actions of G on transitive sets. So suppose G acts transitively on S. And let H be the fixed group of the group fixing a little point S. And what we want to do is how can we reconstruct S from just from H and G? So I suppose we're given a group G in the subgroup H fixing a point and we somehow forgot what S is. Can we reconstruct S just from H? And that's quite easy because any element S in S is, sorry, we pick some point S such that H is the things fixing it. And now any other point T is of the form G times S for some G. But different values of G might give the same point T. So suppose T is equal to G1S and T is equal to G2S. Well, that means G2, G1, G2 to the minus G1S equals S. So G2 to the minus one G1 is in H. So G1 is in G2H where this means all elements that are the products of G2 by something at H. This is called a left coset of H. It's where you take all elements of H and multiply by them the same element of G. And you can see this is the same as saying that G1H is equal to G2H. So this means you can actually identify the elements S with left cosets of H. These correspond to left cosets of H and the correspondence is given as follows. If we've got an element T of S, this corresponds to the left coset of all elements G. So we take the set of all G such that G of S equals T. And as we've more or less seen above, this set is a left coset. On the other hand, if we've got a left coset of the form GH, we can map this to the element G of S. And this doesn't depend on which element of this coset we take. If we take a different element of the coset and apply it to S, then we get the same element of S. So this gives the geometric interpretation of cosets. The left cosets can be thought of as just the elements of a set acted on by G. By the way, we're talking about left cosets, which suggests what is a right coset. Well, a right coset is much the same except you multiply all elements of H on the right by G. What do they correspond to? Well, left cosets you get when G is acting on the left on a set. And right cosets you get when G is acting on the right on a set. And this is incredibly confusing when you start talking about left actions of a G on a set and right actions of G on a set. So we just stick with left actions and left cosets and forget about right actions and right cosets. So this suggests how to reconstruct S from G. So given H and G, reconstruct a set S acted on by G as S is equal to set of left cosets of H. And what we want to do is to just check that this gives us a nice set on which G acts. And this is not very difficult. And we just have to do a lot of rather small checks. First of all, any two cosets G1H, G2H are the same or disjoint. And that's easy to check. If they're not disjoint, they've got an element in common. So suppose they've got the same element in common, that means that G1H1 is equal to G2H2 for some G1 and G2. This is assuming G1 of H equals G2 of H. Well, that means that G1 is equal to G2 times H3, where H3 is H2 times the inverse of H. And this means that G1 of H is just equal to G2H3 of H. And H3 of H is just H. So this is equal to G2 of H. So if two cosets have a single element in common, then they're actually the same. So G is a disjoint union of the cosets GH. And now G acts on the cosets by saying G of G1H is just the cosets G1 of H. So this is just a left coset of H by the element G1. And you can check this as a group action and checking it's a group action is kind of trivial and not very interesting to watch, so I won't bother doing it. You've just got to check the rules for a group action, which is one line calculation. So we've shown that if we've got an action of G on a set, then we get subgroups from it. And conversely, we've shown that if we got any subgroup of a group, then it comes from the action of that group on a set of cosets. So how do we use this for Lagrange's theorem? Well, we also notice that any two cosets have the same size. So all we have to do is to show that a coset GH has the same size as H, which is the coset 1H, because then any two cosets were the same size as this. And that's very easy because if we've got a set GH and a set H, we can map any element T of this to G to the minus 1T, which will be an H. And conversely, if we've got an element H here, we can just map it to G of H. And you can check that these are inverse maps from this coset to H. Notice that this uses inverses of the group. If we had omitted the existence of inverses in the axioms for a group, we'd get something called a semi-group, a typical example of a semi-group might be the positive integers. And for semi-groups, you can still define cosets, but they don't necessarily have the same size. Well, now we get Lagrange's theorem. G is a disjoint union of the cosets of H. Secondly, all cosets are the same size. And from these two facts, this implies the order of G, which you remember we write like that, is equal to the order of H times the number of cosets. So the order of H divides the order of G. You sometimes denote that a number divides another number by putting a vertical line, but that's not terribly good notation because then we would write H divides the order of G, which would be horribly ambiguous. So let's not do that. So if that gives the proof of Lagrange's theorem, it has a very useful corollary. You remember that the order of G is the order of H times the number of cosets. And this is a very useful way for calculating the order of G. For instance, we can do this for an icosahedron. In fact, we did it for an icosahedron earlier, but I'll just remind you again. Suppose we want the order of an icosahedron. We fix the group of symmetries of an icosahedron. We fix a point. We might fix a vertex, say. And we consider the subgroup H fixing that vertex. Well, that's going to have order five. So the order of G is the order of that subgroup, which is five times the number of cosets. And the number of cosets is just the number of other vertices because you remember cosets correspond to things acted on by G. So the order of symmetries of the icosahedron is five, the number of rotations fixing a point, times the number of vertices, which is 60. Alternatively, we could use the group, the subgroup fixing a triangle, meaning mapping the triangle itself. And this is order three. And there are 20 cosets corresponding to the 20 triangles of the icosahedron. So we get the order is three times 20, which is 60. And we can do the same thing with edges, where the subgroup fixing an edge has ordered two and there are 30 edges. So the order of the group is two times 30, which is 60. So in general, if you've got an unknown group and you want to know its order, you try and find the subgroup whose order you know and count the number of cosets by trying to identify the cosets of H with some geometric thing acted on by G. There are two applications of Lagrange's theorem that I will mention to finish off this lecture. The first is Fermat's theorem, not to be confused with Fermat's last theorem. This is the theorem that says that x to the p is congruent to x mod p, meaning x to the p minus x is divisible by p, where the p is, of course, prime. And how do we get this from Lagrange's theorem? Well, Lagrange's theorem says that the order implies that the order of an element G of a group G divides the order of G. And the reason is that the order of G is the order of a subgroup, and the order of a subgroup divides the order of G. So this implies that little G to the order of G is equal to 1 because this is a multiple of the order of G and G to the power of the order of G is 1. And we will now show this implies Fermat's theorem. What we do is we take the group G to be Z over PZ star. This means the non-zero elements of Z modulo PZ. And by number theory, if something is not the zero element of this set, then it has an inverse, because if it's non-zero, it's co-prime to P. So if the element a in here is non-zero, it's co-prime to P. So we can solve ax plus bp equals 1 using Euler's theorem. And this means that x is the inverse of a modulo P. So we've got a group of order P minus 1. So if x is not zero in Z modulo PZ, then x to the P minus 1 is common to 1 mod P by this theorem here. We're just taking x to the power of the order of the group. And now that applies to all non-zero elements x. And if you multiply both sides by x, you're now getting an equation that's also true for x equals zero. So this holds for all x. So for example, 2 to the 10 minus 2 is divisible by 11. And you can write down all sorts of congruences like that that aren't immediately obvious. There's an extension of this due to Euler. So Fermat's theorem applies to primes. Euler's theorem says that if x is co-prime to some number m greater than zero, then x to the phi of m is common to 1 mod m. Well, what's phi of m? Well, I'll explain in a moment. What we do here is we look at the group Z over mz star. This is all integers mod m that are co-prime to m. And this forms a group by exactly the same argument we had for when m was a prime. That if something is co-prime to m, then we can find an inverse. So if this group is called g, then it means x to the g is common to 1 mod m if x is in g. In other words, if x is co-prime to m. Well, what is the order of g? Well, the order of g is equal to phi of m because phi of m is defined to be the order of g. So this is Euler's phi function. So let's just see an example of this. Let's take m equals 12. Then z over 12z star has four elements, 1, 5, 7, and 11 because these are the four numbers modulo 12 that are co-prime to 12. So phi of 12 is equal to 4. So this says 1 to the 4 is common to 1 mod 12 and 5 to the 4 is common to 1 and 7 to the 4 is common to 1 and 11 to the 4 is common to 1. Actually, this is a pretty bad result because, in fact, 5 squared is common to 1, 7 squared is common to 1, and 11 squared is common to 1. So Euler's theorem is a bit sloppy because it just says that any element to the power of 4 is 1 mod 12 but, in fact, any element squared is 1 mod 12. And, in fact, Euler's theorem for most non-prime moduli is not the best possible. Okay, next lecture we will move on to groups of order 4 and discuss things like products of groups.